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Say your trying to find the Thevenin equivalent of a linear circuit with dependent and independent sources.

You would find \$V_{OC}\$ (voltage across the short circuit) and \$I_{SC}\$, the source current, and the Thevenin resistance would be \$R_{Th} = \frac{V_{OC}}{I_{SC}}\$.

But is it possibly for both the voltage across the short circuit and the source current to be zero, leaving the Thevenin resistance indeterminate? Why or why not? What about the source current being zero?

Please note I am specifically talking about a circuit containing dependent and independent sources.

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  • \$\begingroup\$ Where you say "source current" I think you want to say "short-circuit current". \$\endgroup\$ – The Photon Apr 13 '14 at 3:25
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It's entirely possible for both the short circuit current and open-circuit voltage to be zero. For example, if the network consists of a single resistor (or probably any linear network with no independent sources).

In this case you just need to use a different pair of "test points" to find the equivalent resistance. For example, you can use V=0 (the short-circuit case) and V = 1 V. Then \$R_{th}=\frac{1}{I(V=1) - I(V=0)}\$, or more generally \$R_{th}=\frac{V_1-V_2}{I(V_1) - I(V_2)}\$.

Basically one way to state Thevenin's theorem is that the I-V characteristic of any one-terminal network of independent sources and linear elements will be a straight line. And \$R_{th}\$ is the slope of that line, \$\dfrac{\mathrm{d}v}{\mathrm{d}i}\$. Just as with any other linear function, you can can use any two points to find the slope of the line as rise/run.

Edit

You would find \$V_{OC}\$ (voltage across the short circuit) and \$I_{SC}\$, the source current, and the Thevenin resistance would be \$R_{Th} = \frac{V_{OC}}{I_{SC}}\$.

You've mis-stated this a bit (where I added emphasis).

The usual way to find the Thevenin or Norton equivalent circuit is to find \$V_{OC}\$, the voltage across an open-circuit on the output, and \$I_{SC}\$, the current through a short-circuit on the output.

It happens that \$V_{OC}\$ is also the Thevenin-equivalent voltage source value and \$I_{SC}\$ is the Norton-equivalent current source value. But when you're measuring an arbitrary network to find the equivalents, you should think of these as the open-circuit output voltage and short-circuit output current.

But is it possibly for both the voltage across the [open] circuit and the [short-circuit] current to be zero, leaving the Thevenin resistance indeterminate?

I believe \$V_{OC}\$ and \$I_{SC}\$ will both be zero whenever there are no non-zero independent sources in the network being modeled. However this does not mean that the Thevenin resistance is indeterminate, as I explained above.

You could also contrive a case where there's a non-zero independent source and a dependent source that directly opposes it, like for example,

schematic

simulate this circuit – Schematic created using CircuitLab

Can circuits with dependent and independent have indeterminate Thevenin resistances?

I think this is a different question. The Thevenin equivalent resistance will only be indeterminate if it goes to infinity, for example when the network being modeled is an ideal current source.

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  • \$\begingroup\$ But I specifically asked about circuits with independent sources. Can circuits with dependent and independent have indeterminate Thevenin resistances? \$\endgroup\$ – dfg Apr 13 '14 at 3:12
  • \$\begingroup\$ That is the point of my last paragraph. Even with linear independent sources, the I-V characteristic is still a straight line, and you can use any two points to find the slope of the line. \$\endgroup\$ – The Photon Apr 13 '14 at 3:21

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